practice problem 1
- Determine the drag coefficient of a 75 kg skydiver with a projected area of 0.33 m2 and a terminal velocity of 60 m/s.
- By how much would the skydiver need to reduce her project area so as to double her terminal velocity? How would she accomplish this?
Terminal velocity for a falling object occurs when the drag on the object equals its weight.
R = W ½ρCAvt2 = mg
Solve for projected area, substitute values, and compute. (The density of air is in this book somewhere.)
C = 2mg ρAv2 C = 2(75 kg)(9.8 m/s2) (1.207 kg/m3)(0.33 m2)(55 m/s)2 C = 1.22
This agrees with the range of values stated in the table on the discussion page of this topic.
We still start with the principle that drag equals weight.
R = W ½ρCAvt2 = mg
But this time we'll solve for the terminal velocity instead of the drag coefficient.
vt = √ 2mg ρCA
Don't plug in any numbers, just look at the way terminal velocity is related to projected area. Projected area is in the denominator, under a radical sign. That means terminal velocity is inversely proportional to the square root of projected area. That means the skydiver would have to reduce her projected area to one-quarter of its original value.
2vt ∝ √ 1 ⇐ vt ∝ √ 1 ¼A A
The skydiver can do this by changing her orientation from horizontal to vertical — from spread eagle to head first.
practice problem 2
Did Monsieur Mariotte make a good prediction? Keep in mind that he had no way to test it. The stopwatch was two centuries away at least. I converted the pre-revolution units to metric ones for you. Pieds are the French equivalent of and very similar to English feet. Lignes, lines, are like inches.
Voici des Tables faites sur cette hypothèse, par lesquelles on connoîtra combien une balle de plomb de six lignes de diamètre passera de pieds en chaque seconde en descendant; combien elle en passera dans tel nombre de secondes qu'on voudra choisir; quand elle cessera d'accélérer son mouvement; quelle sera sa vites se complette; & combien elle parcourra de pieds avant que de l'acquérir.
These tables, made by applying this hypothesis, show how many feet a lead ball six lines [1.3536 cm] in diameter will fall in each second; how many feet it will fall in any number of seconds we choose; when and where it stops accelerating; what will be its final speed; & how many feet it will cover before acquiring it.
Adapted from Edme Mariotte, 1673
- Construct a graph of distance vs. time from Mariotte's predicted values and use it to determine the terminal velocity of his hypothetical lead ball.
- Use the contemporary drag equation, R = ½ρCAv2, that evolved from Mariotte's hypothesis to determine the terminal velocity of his hypothetical lead ball.
- How do the results of your previous two analyses compare?
Once this is graphed, you can kind of see what's going on. The last five data points line up pretty nearly perfectly. Make a line of best fit using those last five points. This part of the problem can be solved in a few different ways. I chose this method since it gives a reasonable answer with a minimal amount of effort.
Since speed is the rate of change of distance with time, the slope of the line added is the terminal velocity.
vmariotte = 44.76 m/s
This part of the problem can also be solved in a few different ways. Let's use as much algebra with symbols as we can. I chose this method since it involves the fewest number of calculations.
At terminal velocity, drag equals weight.
R = W
Replace those two symbols with appropriate contemporary equations.
12ρCAv2 = mg
Add subscripts to identify the two very different materials.
12ρairCAv2 = mleadg
Relate the mass of the lead sphere to its density
mlead = ρleadV
Recall the equations for the projected area of a sphere (the area of a circle) and the volume of a sphere.
A = πr2 V = 43πr3
Now do a triple substitution.
12ρairCπr2v2 = ρlead43πr3g
Simplify a bit.
12ρairCv2 = ρlead43rg
Solve for speed.
v = √ 8ρleadrg 3ρairC
List the quantities with numbers and units.
ρlead = 11,350 kg/m3 r = 0.006768 m, half the diameter given and then converted to meters g = 9.8 m/s2, or use standard gravity 9.80665 m/s2 if you wish ρair = 1.207 kg/m3 at 20 °C, but anything close to this is good enough C = 0.5 and no units
Put them into the equation.
v = √ 8(11,350 kg/m3)(0.006768 m)(9.8 m/s3) 3(1.207 kg/m3)(0.5)
Compute the answer.
vmodern = 57.67 m/s
The two answers we got are of the same order of magnitude — the minimum amount one needs to be considered "right" in physics.
44.76 m/s ≈ 57.67 m/s
I thought I might have more to say on this, but I don't. Given the state of development of fluid mechanics at the time (it was just being invented) I'd say Mariotte basically got it right. Your answer is allowed to differ only slightly from mine — let's say, to within an order of magnitude.
practice problem 3
- drag is directly porpotional to speed and
- drag is proportional to the square of speed.
Start with Newton's second law of motion. Identify the relevant forces. Weight (W) pulls the skydiver down. Drag (R) pushes her back up. We'll make down be the positive direction, since that's where the skydiver's headed.
∑F = ma W − R = ma mg − bv = ma
Acceleration is the rate of change of velocity with time.
mg − bv = m dv dt
This is a first order differential equation — equation because there's an equals sign, differential because it contains two infinitesimals (dt and dv), and first order because the infinitesimals are not raised to anything higher than the first power. This kind of equation is best solved by separation of variables. Place all the time terms on one side (a rather lonely side) and all the velocity terms on the other side (a somewhat crowded side).
dt = m dv mg − bv
Integrate both sides. Time starts at zero and ends at some later time. Velocity starts at zero and ends at some later velocity. Note that the symbols t and v are doing double duty in the equation. They are acting as both variables and upper limits.
dt = m dv mg − bv
Complete the integral…
t = − m ⎡
ln|mg − bv| b
Evaluate over the limits…
(t − 0) = − m (ln |mg − bv| − ln |mg|) b
− b t = ln mg − bv m mg − b t = ln 1 − b v m mg
Remember, we're trying to make velocity into a function of time. We need to get v out of the logarithm. Raise both sides of the equation into a power of e.
e−bt/m = 1 − b v mg
Finish things off with a little bit of algebra…
v = mg (1 − e−bt/m) b
Let's check the limits of this equation to see if they make sense. This function returns the value v = 0 when t = 0.
v(0) = mg (1 − e−b0/m) b v(0) = mg (1 − 1) = 0 b v(0) = 0
This agrees with our condition that the skydiver wasn't moving at first.
This function always increases, but it never quite reaches a final value. It approaches v = mg/b as we get closer and closer to t = ∞.
v(∞) = mg (1 − e−b∞/m) b v(∞) = mg (1 − 0) b v(∞) = mg b
This is our terminal velocity. We'd get the same thing if we set drag equal to weight and solved algebraically for speed.
One last test. What happens to our function if we let b = 0? What happens if we get rid of drag?
v(b = 0) = mg (1 − e−0t/m) = 0 v(b = 0) = mg (1 − 1) = mg 0 v(b = 0) = mg 0 = uh oh! 0
What is the limit of zero divided by zero? To answer that question, we'll use a little trick called L'Hôpital's rule — named for the French mathematician Guillaume de l'Hôpital (1661–1704). The ratio of two limits that approach zero is the same as the ratio of the limits of their first derivatives. (There's a bit more to this rule, but I'll leave it to your math teacher to fill in the details.)
lim f(x) = lim df/dx g(x) dg/dx
For our problem, where the limiting variable is b, we'll let…
f(b) = (1 − e−bt/m) ⇒ df = t e−bt/m db m g(b) = b ⇒ dg = 1 db
So now we need to take the limit of this instead…
v(b = 0) = mg t e−bt/m 1 m v(b = 0) = gt v(b = 0) = oh yeah!
When we get rid of drag, we get back the velocity-time equation for uniform acceleration — in other words, free fall…
v = gt
The solution makes sense.
Repeat the previous approach using a drag that is proportional to speed squared.
∑F = ma W − R = ma mg − bv2 = ma
Rearrange into a first order differential equation…
mg − bv2 = m dv dt
Separate the variables…
dt = m dv mg − bv2
Integrate both sides.
dt = m dv mg − bv2
Let me be honest here. I have no idea what to do next. I had to consult an expert. Apparently, the inetgral on the right works out to…
t = √ m tanh−1 ⎛
v √ b ⎞
Make velocity the subject.
v = √ mg tanh ⎛
t √ bg ⎞
Have you ever seen anything that looks like this before? Did you notice the odd function? That is not a typo. I did not accidentally drop an "h" into "tan" (short for tangent). I really meant to write "tanh". Let me digress briefly.
The normal trig functions (sine, cosine, tangent, etc.) are ratios of the sides of a right triangle. A standard way to imagine this triangle is with one vertex at the center of a circle. In this configuration, the hypotenuse of our standard triangle is the radius of the circle in which the triangle is embedded. Take that radius and sweep it around the circumference. Watch the ratios evolve. This is the traditional way to graph them —as functions of an angle swept around a circle.
Well now, who says we have to do it this way? Who says we have to use a circle? Well now, I would if you asked me, since I tend to relate angles with fractions of a circle, but I'm just a physicist. Mathematicians on the other hand are some clever people. One of them actually had the temerity to respond to the question, "Why a circle?" with another, more interesting question, "Why not a hyperbola?" And the rest was history.
There are trig functions defined on a circle and trig functions defined on a hyperbola. The former are called sine (sin), cosine (cos), tangent (tan), and so on (cot, sec, and csc). The latter are called hyperbolic sine (sinh), hyperbolic cosine (cosh), hyperbolic tangent (tanh), and hyperbolic so on (coth, sech, and csch). Here's what the mathematicians have determined these functions look like (where x is the "angle" measured in radians).
sinh (x) = e+x − e−x 2 cosh (x) = e+x + e−x 2 tanh (x) = e+x − e−x e+x + e−x
So what does that mean for us? Make the following replacement in the definition of tanh.
x = t √ bg m
Look at the big beautiful pile of symbols we get.
e + t √ bg − e − t √ bg ⎞
v = √ mg m m b e + t √ bg + e − t √ bg m m
Isn't that something? Astoundingly, this little mathematical diversion turns out to be useful. Hyperbolic trig functions are more than just an intellectual challenge. They wind up being the solution to some real world problems.
Let's not test the limits of this equation. There are so many terms to keep track of. It wouldn't be much fun. Instead, let's taste and compare. Graph the two solutions side by side.
Both solutions are exponential at their core. The speed squared model approaches its terminal velocity faster than the directly proportional model. This makes sense, since squaring the speed makes the drag increase more quickly. Whether the terminal velocities would differ is another matter. The graphs above assume that both models generate the same constant of proportionality. Since these are entirely hypothetical models, no real comparison can be made.
practice problem 4
This tab-delimited text file consists of just the power and top speed data for 122 cars tested by Road & Track (paid link) magazine in 1998. Use the data in this file and your favorite analysis software to determine the model that best describes aerodynamic drag for automobiles; that is, determine the value of the power n in the generalized drag equation…
R = −bvn
My favorite data analysis software gave me the following coefficients when I graphed P vs. vmax and did a power curve fit.
It looks like aerodynamic drag for cars is proportional to the square of speed. Bernoulli's equation tells us that drag is proportional to the square of speed and I see a power that's approximately 2 in the curve fit above. Isn't life grand when everything behaves exactly as you expect it to.
Halt! Proceed no further with this logic. It is wrong, wrong, wrong, wrong, wrong, wrong, wrong. The graph above shows the relationship between power and speed, not drag force and speed. The question now is how to make the jump from what we have to what we need.
Begin with the definition of work and play around with it a bit.
|P =||W||=||Fs||= Fv|
Replace the nonspecific force F with power law form of aerodynamic drag.
P = Fv = bvn v = bvn + 1
The curve fit that I did gave me one greater than the power in the general relation. This means that the aerodynamic drag on automobiles is proportional to speed not speed squared. Bernoulli's law does not apply for some reason. I don't know how to reconcile this discrepancy. I suspect that automotive engineers use…
R = ½ρCAv2
for drag calculations, not the computationally simple, but physically unrealistic…
R = −bv
but then, I don't know any automotive engineers.